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JEE CHEMISTRY MCQ: HYDROCARBONS

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Analyzing +4 | -1
In the following sequence the product D is,

CH≡CH→HBrA→HBrB→KOH(alc.)C→NaNH2D:\mathrm{CH} \equiv \mathrm{CH}\xrightarrow{{HBr}}A\xrightarrow{{HBr}}B\xrightarrow{{\mathrm{KOH}(\mathrm{alc}\mathrm{.})}}C\xrightarrow{{\mathrm{NaN}{\mathrm{H}_2}}}D:
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Correct option: C
Ethyne
Option C (Ethyne) is correct because the sequence of reactions leads back to the starting material, ethyne.

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